RD Sharma Solutions Class 12 Mathematics Chapter 15 VSA

Edited By Kuldeep Maurya | Updated on Jan 21, 2022 01:03 PM IST

Many students widely use the RD Sharma solution books to clarify their doubts without the help of a teacher or a tutor. Solving the Very Short Answers (VSA) in class 12 mathematics is trickier as the solutions must be small and solved using shortcut methods. However, students can save time and complete the sums effortlessly using this method. And to learn it, the RD Sharma Class 12th VSA book is the best choice.

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RD Sharma Class 12 Solutions Chapter 15 Tangents and Normals - Other Exercise

Chapter 15 - Tangents and Normals Ex 15.1

Chapter 15 - Tangents and Normals Ex 15.2

Chapter 15 - Tangents and Normals Ex 15.3

Chapter 15 -Tangents and Normals Ex-FBQ

Chapter 15 -Tangents and Normals Ex-MCQ

Tangents and Normals Excercise:VSA

Tangents and Normals Exercise Very Short Answer Question 1

Answer :
$\left(x_{1}, y_{1}\right)=(1,2)$
Hint:
Use the slope of the tangent.
Given:
Here the curve $y=x^{2}-2 x+3$ , where the tangent is parallel to $x -$ axis.
We have to find the point on the given curve.
Solution :
The slope at the $x -$ axis is $0$ .
Now,
Let $\left ( x_{1},y_{1} \right )$ be the required point.
Since the point lies on the curve.
Hence $y_{1}=x_{1}^{2}-2 x_{1}+3$ ....(i)
Then, $y=x^{2}-2 x+3$
$\Rightarrow \quad \frac{d y}{d x}=2 x-2$
Slope of the tangent at $(x, y)=\left(\frac{d y}{d x}\right)_{\left(x_{1}, y_{1}\right)}$
$= 2x_{1}-2$
Here $\frac{d y}{d x}=0$
$\begin{aligned} &\Rightarrow \quad 2 x_{1}-2=0 \\ &\Rightarrow \quad x_{1}=1 \end{aligned}$
From equation (i)
$\begin{aligned} &\Rightarrow \quad y_{1}=1-2+3=2 \\ &\Rightarrow \quad y_{1}=2 \end{aligned}$
Hence required point is $\left ( x_{1},y_{1} \right )= (1,2)$

Tangents and Normals Exercise Very Short Answer Question 2

Answer : Slope of tangent at $t=2 \text { is } \frac{6}{7}$
Hint : Use equation of tangent,
$\frac{y-y_{1}}{x-x_{1}}=\left(\frac{d y}{d x}\right)_{\left(x_{1}, y_{1}\right)}$
Given :
Here, $x=t^{2}+3 t-8 \text { and } y=2 t^{2}-2 t-5$
We have to find the slope of tangent to given curve at $t = 2$
Solution :
The equation of the given curve is
$\begin{aligned} &x=t^{2}+3 t-8 \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; ....(i)\\ &y=2 t^{2}-2 t-5 \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; ...(ii) \end{aligned}$
Differentiating equation (i) with respect to $'t'$
We have,
$\begin{aligned} &\frac{d y}{d t}=4 t-2\\ &\therefore \frac{d y}{d x}=\frac{\frac{d x}{d t}}{\frac{d y}{d t}}=\frac{4 t-2}{2 t+3} \end{aligned}$
Now, slope of tangent at $t=2$
$\left(\frac{d y}{d x}\right)_{t=2}=\frac{8-2}{4+3}=\frac{6}{7}$
Hence, Slope of tangent at $t=2$ is $\frac{6}{7}$ .

Tangents and Normals Exercise Very Short Answer Question 3

Answer : $\frac{dy}{dx}=0$
Hint : The curve is parallel to $x-$ axis then slope of tangent is equal to slope of the $x-$ axis.
Given :
The tangent line at a point $(x,y)$ on the curve $y = f (x)$ is parallel to $x-$ axis.
We have to find the value of $\frac{dy}{dx}$ .
Solution :
Here we have,
Slope of $x-$ axis is equal to zero.
Also, the tangent line at a point $\left ( x,y \right )$ on the curve $y = f (x)$ is parallel to $x-$ axis.
$\therefore$ Slope of tangent $\left ( \frac{dy}{dx} \right )=$ slope of $x-$ axis
$\Rightarrow \frac{dy}{dx}=0$

Tangents and Normals Exercise Very Short Answer Question 4

Answer : $\frac{dy}{dx}=0$
Hint :
1. The slope of y - axis is $\infty$ .
2. Using, $\text{Slope of tangent}=\frac{-1}{\text {slope of normal}}$
Given: The normal of the curve $y=f (x)$ at $(x,y)$ is parallel to y -axis.
We have to write the value of $\frac{dy}{dx}$
Solution:
We know that,
The slope of y - axis is $\infty$
Also, The normal of the curve $y=f(x) \text { at }(x, y)$ is parallel to y - axis.
$\therefore$ slope of normal = slope of y- axis
$\begin{aligned} &\Rightarrow \quad \quad \frac{d y}{d x}=\text { Slope of tangent }=\frac{-1}{\text { slope of normal }} \\ &\; \; \; \; \; \; \; \; \; \; \; =\frac{-1}{\infty}=0 \end{aligned}$
Hence, $\frac{dy}{dx}=0$

Tangents and Normals Exercise Very Short Answer Question 5

Answer : $\frac{dy}{dx}=1\; \text{or} -1$
Hint:
The curve is equally inclined to the axes, then the angle made by the tangent with axes can be $\pm 45^{o}$ .
Given:
Here given that,
The tangent to a curve at a point $(x,y)$ is equally inclined to the co-ordinate axes.
We have to write the value of $\frac{dy}{dx}$ .
Solution :
We know that,
The tangent to a curve at a point $(x,y)$ is equally inclined to the co-ordinate axes, the angle made by the tangent with axes can be $\pm 45^{o}$ .
$\therefore$ $\frac{dy}{dx}= \text {slope of tangent}$
$\Rightarrow \frac{dy}{dx}=\tan (\pm 45^{o})=\pm 1$
Hence, $\frac{dy}{dx}=1$ or $-1$

Tangents and Normals Exercise Very Short Answer Question 6

Answer : $\frac{dx}{dy}=0$
Hint : Slope of the y-axis is $\infty$ .
Given :
Given that
The tangent line at a point $(x,y)$ on the curvey $=f(x)$ is parallel to y-axis.
We have to write the value of $\frac{dx}{dy}$ .
Solution :
We know that,
The slope of the y-axis is $\infty$ .
Also, the tangent line at a point $(x,y)$ on the curvey $=f(x)$ is parallel to y-axis.
$\text {Slope of tangent}=\frac{\text{dy}}{\text{dx}}$ = slope of y-axis
$\begin{aligned} &\Rightarrow \quad \frac{d y}{d x}=\infty \\ &\therefore \frac{d x}{d y}=\frac{1}{\frac{d y}{d x}}=\frac{1}{\infty}=0 \end{aligned}$
Hence, $\frac{dx}{dy}=0$

Tangents and Normals Exercise Very Short Answer Question 7

Answer :
$\text { slope of normal }=\frac{1}{t^{2}}$
Hint : Usingmula,
$\text { slope of normal }=\frac{-1}{\text { slope of tangent }}$
Given :
Here given the curve,
$x=\frac{1}{t} \text { and } y=t$
We have to find the slope of normal.
Solution :
Here,
$x=\frac{1}{t} \text { and } y=t$
Differentiating with respect to $'t'$ , we get
$\begin{gathered} \Rightarrow \quad \frac{d x}{d t}=\frac{-1}{t^{2}} \text { and } \frac{d y}{d t}=1 \\ \therefore \frac{d y}{d x}=\frac{\frac{d y}{d t}}{\frac{d x}{d t}}=\frac{1}{\left(\frac{-1}{t^{2}}\right)} \end{gathered}$
$\Rightarrow \quad \frac{d y}{d x}=-t^{2}$
So, Slope of tangent $=\frac{dy}{dx}= -t^{2}$
$\text { slope of normal }=\frac{-1}{\text { slope of tangent }}$
$=\frac{-1}{-t^{2}}=\frac{1}{t^{2}}$
Hence,
$\text { slope of normal }=\frac{1}{t^{2}}$

Tangents and Normals Exercise Very Short Answer Question 8

Answer : $(x, y)=\left(\frac{1}{4}, \frac{1}{2}\right)$
Hint : $\text { slope of tangent }=\tan \frac{\pi}{4}=1$
Given :
Here the given curve, $y^{2}=x$ where the tangent line makes an angle $\frac{\pi }{4}$ with x- axis.
We have to find the co-ordinate of the point on the given curve.
Solution:
Let the required point be $(x,y)$ .
We know,
The tangent makes an angle $45^{o}$ with the x-axis.
$\therefore$ $\text { slope of tangent }=\tan 45^{\circ}=1$
Since the point lies on the curve
Hence, $y_{1}^{2}=x_{1}$
Now,
$y^{2}=x$
Differentiating with respect to x,
$\begin{array}{ll} \Rightarrow \quad & 2 y \frac{d y}{d x}=1 \\ \Rightarrow & \frac{d y}{d x}=\frac{1}{2 y} \end{array}$
$\therefore \quad \text { slope of tangent }=\left(\frac{d y}{d x}\right)_{\left(x_{1}, y_{1}\right)}=\frac{1}{2 y_{1}}$
$\Rightarrow \quad \frac{1}{2 y_{1}}=1 \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[\because \text { slope of tangent at angle } \frac{\pi}{4}=\tan \frac{\pi}{4}=1\right]$
$\Rightarrow \quad 2 y_{1}=1$
$\Rightarrow \quad y_{1}=\frac{1}{2}$
Now, we have,
$\begin{aligned} &x_{1}=y_{1}^{2} \\ &\Rightarrow x_{1}=\left(\frac{1}{2}\right)^{2} \\ &\Rightarrow \quad x_{1}=\frac{1}{4} \end{aligned}$
Hence, $(x, y)=\left(\frac{1}{4}, \frac{1}{2}\right)$

Tangents and Normals Exercise Very Short Answer Question 9

Answer : $\frac{\pi }{2}$
Hint : Differentiating both equation and find $\frac{dy}{dx}$
Given :
Given that the curve,
$x=e^{t} \cos t \text { and } y=e^{t} \sin t$
We have to find the angle made by the tangent to the given curve.
Solution :
Here,
$x=e^{t} \cos t \text { and } y=e^{t} \sin t$
Differentiating both equation with respect to $'t'$
We have,
$\frac{d x}{d t}=e^{t} \cos t-e^{t} \sin t$
$\begin{aligned} &\text { - }\\ &\begin{aligned} &\frac{d y}{d t}=e^{t} \sin t+e^{t} \cos t \\ &\frac{d y}{d x}=\frac{\frac{d y}{d t}}{\frac{d x}{d t}} \end{aligned}\\ &=\frac{e^{t} \sin t+e^{t} \cos t}{e^{t} \cos t-e^{t} \sin t} \end{aligned}$
Now, slope of tangent $=\left(\frac{d y}{d x}\right)_{t=\frac{\pi}{4}}$
$\begin{aligned} &=\frac{e^{t}\left(\sin \frac{\pi}{4}+\cos \frac{\pi}{4}\right)}{e^{t}\left(\cos \frac{\pi}{4}-\sin \frac{\pi}{4}\right)} \\ &=\frac{\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}}{\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}}}=\frac{\frac{2}{\sqrt{2}}}{0} \\ &\frac{d y}{d x}=\infty \end{aligned}$
Let $\theta$ be the angle made by the tangent with the x-axis.
$\begin{array}{ll} \therefore & \tan \theta=\infty \\ \Rightarrow & \theta=\frac{\pi}{2} \end{array}$
Hence, the angle made by the tangent to the given curve is $\frac{\pi }{2}$ .

Tangents and Normals Exercise Very Short Answer Question 10

Answer : Equation of normal $=2x = \pi$
Hint : Use equation of normal,
$y-y_{1}=\frac{1}{\frac{d y}{d x}}\left(x-x_{1}\right)$
Given :
Here given that the curve
$y=x+\sin x \cos x$
We have to write the equation of normal to the given curve at $x=\frac{\pi }{2}$
Solution:
We have,
$y=x+\sin x \cos x$
On differentiating both sides with respect to $'x'$ , we get
$\frac{d y}{d x}=1+\cos ^{2} x-\sin ^{2} x$
Now we know that,
$\begin{aligned} &\text { slope of tangent }=\left(\frac{d y}{d x}\right)_{x=\frac{\pi}{2}} \\ &=1+\cos ^{2}\left(\frac{\pi}{2}\right)-\sin ^{2}\left(\frac{\pi}{2}\right) \\ &=1+0-1 \\ &=0 \end{aligned}$
Now,
When $x=\frac{\pi}{2} \Rightarrow y=\frac{\pi}{2}+\sin \frac{\pi}{2} \cos \frac{\pi}{2}$
$\Rightarrow y=\frac{\pi }{2}$
$\therefore \quad\left(x_{1}, y_{1}\right)=\left(\frac{\pi}{2}, \frac{\pi}{2}\right)$
So, equation of normal
$\begin{aligned} &y-y_{1}=\frac{-1}{\text { slope of tangent }}\left(x-x_{1}\right) \\ &y-\frac{\pi}{2}=\frac{-1}{0}\left(x-\frac{\pi}{2}\right) \end{aligned}$
$\begin{array}{ll} \Rightarrow \quad & x=\frac{\pi}{2} \\ \Rightarrow & 2 x=\pi \end{array}$
Hence the equation of the given curve at $x=\frac{\pi}{2} \text { is } 2 x=\pi$ .

Tangents and Normals Exercise Very short Answers Question 11

Answer : $\left(x_{1}, y_{1}\right)=\left(\frac{1}{2}, 1\right)$
Hint :
For line, the slope of a line is denoted by m.
$y=m x+c$ , where m is slope of line.
Given :
Given that the curve, $y^{2}=3-4 x$
Where the tangent parallel to the line, $2 x+y-2=0$
We have to find the co-ordinate of the point on the given curve.
Solution :
Let $\left(x_{1}, y_{1}\right)$ be the required point
We know that,
If $2 x+y-2=0$
$\Rightarrow \quad y=-2 x+2$
Comparing with equation:
$y=mx+c$
We get, $m=-2$
Slope of the given line $=-2$
Since, the point lies on the curve
So $y_{1}^{2}=3-4x_{1}$ ...(i)
Now, $y^{2}=3-4x$
Differentiate with respect to x
$\begin{array}{ll} \Rightarrow \quad & 2 y \frac{d y}{d x}=-4 \\ \therefore & \frac{d y}{d x}=\frac{-2}{y} \end{array}$
Slope of tangent $=\left(\frac{d y}{d x}\right)_{\left(x_{1}, y_{1}\right)}=\frac{-2}{y_{1}}$
Here, slope of tangent=slope of the line
$\begin{aligned} &\Rightarrow \quad \frac{-2}{y_{1}}=-2 \\ &\Rightarrow \quad y_{1}=1 \end{aligned}$
From equation (i), we get
$\begin{aligned} & 1=3-4 x_{1} \\ \Rightarrow & 4 x_{1}=2 \\ \Rightarrow & x_{1}=\frac{1}{2} \end{aligned}$
Hence, $\left(x_{1}, y_{1}\right)=\left(\frac{1}{2}, 1\right)$

Tangents and Normals Exercise Very short Answers Question 12

Answer : Equation of tangent, $x+y=2$
Hint: Use the equation of tangent,
$y-y_{1}=\frac{d y}{d x}\left(x-x_{1}\right)$
Given : Here the curve,
$y=x^{2}-x+2$ at the point where it crosses the y-axis.
We have to write the equation of the tangent.
Solution :
We know that,
When the curve passes through y - axis, then the point on the curve is of them $(0,y)$
Now, $y=x^{2}-x+2$
Differentiating with respect to x,

$\frac{d y}{d x}=2 x-1$
Slope of tangent $=\left(\frac{d y}{d x}\right)_{(0,2)}=2(0)-1$
$=-1= m (say)$
$\therefore \quad\left(x_{1}, y_{1}\right)=(0,2)_{\text {and }} m=1$
Equation of tangent,
$\begin{aligned} & y-y_{1}=m\left(x-x_{1}\right) \\ \Rightarrow & y-2=-1(x-0) \\ \Rightarrow & x+y=2 \end{aligned}$
Hence the required equation of tangent, $x+y = 2$ .

Tangents and Normals Exercise Very short Answers Question 13

Answer : Angle = $0^{o}$
Hint : First, find the slope of given curves then use themula:
$\tan \theta=\left|\frac{m_{1}-m_{2}}{1+m_{1} m_{2}}\right|$
Given :
Given curve,
$\begin{aligned} &y^{2}=4 x_{\text {and }} \\ &x^{2}=2 y-3 \end{aligned}$
We have to find the angle between the given curves.
Solution :
Given :
$y^{2}=4x$ ...(i)
$x^{2}=2y-3$ ...(ii)
On differentiating equation (i) with respect to x, we get
$2y\frac{dy}{dx}=4$
$\Rightarrow \; \; \; \; \; \frac{dy}{dx}=\frac{2}{y}$
$\begin{array}{ll} \Rightarrow & m_{1}=\left(\frac{d y}{d x}\right)_{(1,2)}=\frac{2}{2} \\ \Rightarrow \quad & m_{1}=1 \end{array}$
Now, differentiating equation (ii) with respect to x, we get
$2 x=2 \frac{d y}{d x}$
$\Rightarrow \quad \frac{d y}{d x}=x$
$m_{2}=\left(\frac{d y}{d x}\right)_{(1,2)}=1$
$\Rightarrow \quad m_{2}=1$
As we know that,
$\begin{aligned} &\quad \tan \theta=\left|\frac{m_{1}-m_{2}}{1+m_{1} m_{2}}\right| \\ &\Rightarrow \quad \tan \theta=\left|\frac{1-1}{1+1}\right| \\ &\Rightarrow \quad \tan \theta=0 \\ &\Rightarrow \quad \theta=0^{\circ} \end{aligned}$
Hence the required angle between the curve is $0^{o}$

Tangents and Normals Exercise Very short Answers Question 14

Answer : Required angle $=\frac{\pi }{2}$
Hint :

If $m_{1}=m_{2}$ , they are parallel.
If $m_{1} m_{2}=-1$ , they are perpendicular to each other.

Given :
Given that the curve,
$y=e^{-x} \text { and } y=e^{x}$
We have to find the angle between the given curves at the point of intersection.
Solution :
Given,
$y=e^{-x}$ ....(i)
$y=e^{x}$ .....(ii)
From (i) and (ii), we get
$\begin{aligned} & e^{x}=e^{-x} \\ \Rightarrow & x=0 \end{aligned}$
Substituting the value of x in equation (ii), we get
$y=1$
So, the point of intersection of the two curves is $(0,1)$
On differentiating (i) with respect to x, we get
$\begin{aligned} &\frac{d y}{d x}=e^{x} \\ &m_{2}=\left(\frac{d y}{d x}\right)_{(0,1)}=1 \end{aligned}$
$\begin{aligned} & \therefore & & m_{1} m_{2} &=-1 \\ \text { Since } & & m_{1} m_{2} &=-1 \end{aligned},$
Hence, they are perpendicular to each other.
Hence, the required angle $=\frac{\pi }{2}$

Tangents and Normals Exercise Very short Answers Question 15

Answer : Slope of normal = 9
Hint :
$\begin{aligned} &\text { slope of normal }=\frac{-1}{\text { slope of } \text { tan gent }} \\ &\text { i.e., } \frac{-1}{\frac{d y}{d x}} \end{aligned}$
Given :
Given curve,
$y=\frac{1}{x}$
We have to write the slope of normal to the given curve at point $\left(3, \frac{1}{3}\right)$
Solution :
Here,
$y=\frac{1}{x}$
On differentiating with respect to x, we get
$\frac{d y}{d x}=\frac{-1}{x^{2}}$
Now,
$\begin{aligned} &\text { slope of } \text { tan gent }=\left(\frac{d y}{d x}\right)_{(3, \frac{1}{3})}=\frac{-1}{9} \\ &\text { slope of normal }=\frac{-1}{\text { slope of tan gent }} \end{aligned}$
$=\frac{-1}{\left(\frac{-1}{9}\right)}=9$
Hence, the Slope of normal = 9

Tangent and Normals Exercise Very short Answers Question 17

Answer:

Hint :
$\begin{aligned} &\text { slope of normal }=\frac{-1}{\text { slope of } \text { tangent }} \\ &\text { i.e., } \frac{-1}{\frac{d y}{d x}} \end{aligned}$
Given :
Given curve,
$y=cosx \ \ \ \text{ at } (0,1)$
We have to write the slope of normal to the given curve at point $(0, 1)$
Solution :
Here,
$y=cosx \ \ \ \text{ at } (0,1)$
On differentiating with respect to x, we get
$\frac{d y}{d x}=-sinx$
Now,
$\begin{aligned} &\text { slope of } \text { tangent }=\left(\frac{d y}{d x}\right)_{(0,1)}=0 \\ &\text { slope of normal }=\frac{-1}{\text { slope of tangent }} \end{aligned}$
$=\frac{-1}{0}=\infty$
Hence, normal line is x=0.

Tangent and Normals Exercise Very short Answers Question 18

Answer:

Answer : $y=x$ is required equation of tangent
Hint :
$\text { Equation of tangent }=\left(y-y_{1}\right)=\frac{d y}{d x}\left(x-x_{1}\right)$
Given :
Given curve,
$y=\sin x$
We have to write the equation of tangent drawn to the given curve at point (0,0).
Solution:
$y=\sin x$
Differentiating both sides with respect to x, we get
$\begin{aligned} &\frac{d y}{d x}=\cos x \\ &\left(\frac{d y}{d x}\right)_{(0,0)}=\cos 0=1 \end{aligned}$
Hence the equation of tangent at $(0,0)$ ,
$\begin{aligned} &(y-0)=1(x-0) \\ &y=x \end{aligned}$
Hence, $y=x$ is the required equation.

Tangent and Normals Exercise Very short Answers Question 18

Answer:

Tangent and Normals Exercise Very short Answers Question 19

Answer : $0$
Hint :
Simply we will find $\frac{dy}{dx}$ at $x = \frac{\pi }{6}$
Given:
Given curve,
$y=2 \sin ^{2} 3 x$
We have to find the slope of the tangent to the given curve at $x = \frac{\pi }{6}$
Solution :
$y=2 \sin ^{2} 3 x$
Differentiating both sides with respect to x, we get
$\begin{aligned} &\frac{d y}{d x}=2 \times 2 \sin (3 x) \times \cos (3 x) \times 3 \\ &\Rightarrow \quad\left(\frac{d y}{d x}\right)_{x=\frac{\pi}{6}}=12 \times \sin \left(\frac{\pi}{2}\right) \times \cos \left(\frac{\pi}{2}\right) \end{aligned}$
$=12\times1\times0$
$\left(\frac{d y}{d x}\right)=0$
Hence the slope of tangent = 0

Chapter 15 in mathematics, Tangents, and Normals consists of three exercises, ex 15.1, ex 15.2, and ex 15.3. Class 12 RD Sharma Chapter 15 VSA Solution includes concepts like finding the point of the curve, finding the slope of the tangent, slope of normal at a point, equation of the normal, coordinates of the point, and angles between the curve. RD Sharma Class 12 Solutions Tangents and Normals VSA are around 19 questions given in this exercise. Students can use the RD Sharma Class 12 Chapter 15 VSA solution book for reference.

It is mainly recommended by the CBSE board schools to its students, as the RD Sharma solutions follow the NCERT syllabus. Students can work out the practice questions to understand the steps clearly. The RD Sharma Class 12th VSA book answers are given in simple, easy methods to solve the Very Short Answers quickly. It helps the student use minimal time to solve this section during examinations.

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